Question

In the given figure, $AB=AC$ and $\angle DBC=\angle ECB={90}^{\circ }$,then $BD=CE$ .

• A. true
• B. false

Answer 1

In triangle ABC,
$AB=AC$ thus, $\angle ABC=\angle ACB$ (I) (Opposite angles of equal sides of triangle are equal)
Now, in $△$s DBC and ECB,
$BC=BC$ (Common)
$\angle DBC=\angle ECB=90$ (Given)
$\angle EBC=\angle DCB$ (From I)
Hence, $△DBC\cong △ECB$ (ASA Congruency)
or $BD=CE$ (equal sides of triangle of opposite angle are equal)